Posted by Brianna on Thursday, September 27, 2012 at 2:39pm.
height of ball above cliff is
h = 5 + vy*t - 16t^2
assuming vy is y-component of maximum speed of 86 mph = 126.13 ft/s
h = 0 after .910 sec, so
vy = 9.1 ft/s
now, vy^2 + vx^2 = v^2
9.1^2 + vx^2 = 126.13^2
vx = 125.8 ft/s
ball fell from cliff height
453/125.8 = 3.6 sec
after falling back to cliff height, the distance fallen to ground was
9.1t + 16t^2 = 241 ft
so, the cliff is 241 ft high
Related Questions
College Physics - Bob has just finished climbing a sheer cliff above a beach, ...
physical science - a 2.0 kg rock sits on the edge of a cliff 12m above the beach...
physics - A student stands at the edge of a cliff and throws a stone ...
physics - A student stands at the edge of a cliff and throws a stone ...
Physics - A student stands at the edge of a cliff and throws a stone ...
physics - Multiple part question
.. A student stands at the edge of a cliff...
Physics - A student stands at the edge of a cliff and throws a stone ...
physics - A student stands at the edge of a cliff and throws a stone ...
physics - A rock is thrown from a cliff top at 18m/s, 25 degrees above the ...
physics - A student stands at the edge of a cliff and throws a stone ...
For Further Reading